I came to know from the paper Left Hopf Algebras by Green, Nichols and Taft that one may consider a Hopf algebra whose antipode satisfies only the left (resp. right) antipode condition.
To be more precise, let $\Bbbk$ be a field and $(B,\mu,\eta,\Delta,\varepsilon)$ a $\Bbbk$-bialgebra. We say that $B$ is a left Hopf algebra if there exists a linear endomorphism $S:B\to B$ such that $$S(b_1)b_2=\varepsilon(b)1$$ for every $b\in B$ (i.e. $S$ is a left convolution inverse of the identity morphism).
In Section 3 of Left Hopf Algebras an "artificial" (in my opinion) example of such an object is provided. Are there some more "concrete" or "natural" examples of this construction?
I learn this "obvious" example from Peter Shawenburg. It doesn't give you a direct answer but is very closed.
Take the $A=k\{a,b\}$ free algebra on 2 generators $a$ and $b$. Declare them to be group-like elements, so you have $A=$the semigroup algebra on the free monoid on 2 generators, that is a bialgebra.
Consider the element $ab-1$. It is easy to check that it is skew primitive (in general, the difference of two grouplike is skew primitive), in particular, the ideal generated by it is a coideal. Define $B$= the quotient bialgebra.
Now it is easy to construct examples of non invertible endomorphisms in some (infinite dimensional) vector space with left inverse. (for example, $V=k^{(\mathbb N)}$ and $f(a_1,a_2,\dots,)=(0,a_1,a_2,\dots,)$)
This proves that the element $ba-1$ is not zero in $B:=k\{a,b\}/(ab-1)$.
Notice that $B$ is also a semigroup algebra. The left inverse of $b$ is $a$, but $b$ do not have right inverse. Remark also that in a Hopf algebra, the antipode of a group-like is group-like, so a semigroup algebra is Hopf if and only if the semigroup is a group. This proves that the equation $S(h_1)h_2=\epsilon(h)$ cannot imply $h_1S(h_2)=\epsilon(h)$.